Discussion
The volume of a solid generated by rotating the region between semi-circle \(y = 1 - \sqrt{1 - x^{2}}\) and straight line y = 1, about x axis, is
Answer: A
The volume is \(\pi^{2} - \dfrac{4}{3}\pi\).
The curve \(y = 1 - \sqrt{1 - x^{2}}\) is the lower half of the unit circle centred at (0, 1). The region lies between it and the line \(y = 1\) for \(-1 \le x \le 1\). Rotating about the x axis gives washers of outer radius 1 and inner radius \(1 - \sqrt{1 - x^{2}}\):
\[ \begin{aligned} V &= \pi\int_{-1}^{1}\left[1 - (1 - \sqrt{1 - x^{2}})^{2}\right]dx \\ &= \pi\int_{-1}^{1}\left[2\sqrt{1 - x^{2}} - (1 - x^{2})\right]dx \\ &= \pi\left[2 \cdot \frac{\pi}{2} - \frac{4}{3}\right] \\ &= \pi^{2} - \frac{4}{3}\pi \end{aligned} \]
using \(\int_{-1}^{1}\sqrt{1 - x^{2}}\,dx = \pi/2\) (half the unit circle's area) and \(\int_{-1}^{1}(1 - x^{2})dx = 4/3\). The other options misplace these two terms.
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